demodulating a pam signal generated by sinc- shaped pulses (part 2)
TRANSCRIPT
Demodulating a PAM Signal Generated by Sinc-Shaped Pulses (Part 2)
In Part 1 we determined one way to demodulate a received PAM signal – sampling the signal in the center of the bit period. We assumed that the channel produced no noise or attenuation and that the receiver was perfectly synchronized with the transmitter. Let’s now examine the real-world effects of channel attenuation, noise, and loss of synchronization.
Let’s begin by considering a transmitter using sinc-shaped pulses of amplitude +1 volt and -1 volt, transmitting the data “10010” at 50,000 bits/sec.
10 20 30 40 50 60 70 80 90 100
-2
-1.5
-1
-0.5
0
0.5
1
1.5
sec
volts
1 0 0 1 0
Transmittedsignal
The transmitted waveform is the sum of all the individual pulses (shown by the thick black line)
-2
-1.5
-1
-0.5
0
0.5
1
1.5
0 10 20 30 40 50 60 70 80 90 100
volts
sec
Assuming that the receiver can be perfectly synchronized, one way to demodulate is to sample in the exact center of the received waveform.
This plot assumes no attenuation by the channel and no noise - don’t worry, we will soon be considering these two factors.
If sample is positive, the bit is demodulated as a “1”; if sample is negative, the bit is demodulated as a “0”.
+1V -1V -1V +1V -1V Samples 1 0 0 1 0 Data
Reason #1: There is no intersymbol interference (ISI) in the transmitted waveform in the exact center of each bit period (all interfering pulses have zero value).
There are two good reasons to demodulate by sampling the received waveform in the center of each bit period.
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-2
-1.5
-1
-0.5
0
0.5
1
1.5
sec
volts
1 0 0 1 0
Transmittedsignal
Reason #2: The pulse corresponding to the current bit is at its maximum magnitude in the exact center of its bit period.
There are two good reasons to demodulate by sampling the received waveform in the center of each bit period.
10 20 30 40 50 60 70 80 90 100
-2
-1.5
-1
-0.5
0
0.5
1
1.5
sec
volts
1 0 0 1 0
Transmittedsignal
-2
-1.5
-1
-0.5
0
0.5
1
1.5
10 20 30 40 50 60 70 80 90 100
-0.02 volt at 56 sec
-0.36 volt at 54 sec
-0.69 volt at 52 sec
-1 volt at 50 sec
volts
sec
What happens if the receiver is slightly out of synchronization? Consider the value of the sample for the third bit, which should occur at 50 sec.
Noise Margin is the distance between a noiseless received signal and the threshold at the instant of sampling.
Defining the term Noise Margin
Noise margin is therefore an indication of how strong the noise has to be to “overwhelm” the signal and thereby cause an error.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
10 20 30 40 50 60 70 80 90 100
-0.02 volt at 56 sec
-0.36 volt at 54 sec
-0.69 volt at 52 sec
-1 volt at 50 sec
volts
sec
-2
-1.5
-1
-0.5
0
0.5
1
1.5
10 20 30 40 50 60 70 80 90 100
Noise margin = 0.02 volt at 56 sec
Noise margin = 0.36 volt at 54 sec
Noise margin = 0.69 volt at 52 sec
Noise margin = 1 volt at 50 sec
volts
sec
Assuming no attenuation in the channel,
If the channel produces no attenuation, noise margin is 1 volt at 50 sec, 0.69 volt at 52 sec, 0.36 volt at 54 sec, and 0.02 volt at 56 sec.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
10 20 30 40 50 60 70 80 90 100
Noise margin = 0.01 volt at 56 sec
Noise margin = 0.18 volt at 54 sec
Noise margin = 0.35 volt at 52 sec
Noise margin = 0.5 volt at 50 sec
volts
sec
Assuming 50% attenuation in the channel,
If the channel attenuates the signal so that the received signal has only half the voltage of the transmitted signal, then noise margin is 0.5 volt at 50 msec, 0.345 volt at 52 msec, 0.18 volt at 54 msec, and 0.01 volt at 56 msec.
- 2
- 1 . 5
- 1
- 0 . 5
0
0 . 5
1
1 . 5
1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 1 0 0
v o l t s
s e c
- 2
- 1 . 5
- 1
- 0 . 5
0
0 . 5
1
1 . 5
1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 1 0 0
v o l t s
s e c
Transmitted signal:
Received signal with attenuation and noise:
Attenuation significantly reduces the noise margin, making the received signal much more susceptible to noise. Imprecise (or no) synchronization accentuates the problem.